$\int^{\pi/6}_{0}\sec(2x)\tan(2x)\,dx\, = $
Explanation: Strategy Let's first find the indefinite integral $\int\sec(2x)\tan(2x)\,dx\,$. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\sec(2x)\tan(2x)\,dx\,$, we can use U-substitution. If we let $ {u=2x}$, then ${du=2 \, dx}$ and ${ dx=\dfrac{1}{2}\, du}$. So we have: $\begin{aligned}\int\sec({2x})\tan({2x})\,{dx}\,&=\int\sec( u)\tan( u)\, \cdot {\dfrac12\, du}\,\\\\\\\\ &=\dfrac12\int \sec(u)\tan(u)\, du\,\\\\\\\\ &=\dfrac{1}{2}\cdot \sec(u)+C\\\\\\\\ &=\dfrac12\sec(2x)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{\pi/6}_{0}\sec(2x)\tan(2x)\,dx\, &= \dfrac12\sec(2x)\Bigg|^{\pi/6}_0\\\\\\\\ &=\dfrac{1}{2}\left(\sec\left(\dfrac{\pi}{3}\right)-\sec(0)\right)\\\\\\\\ &=\dfrac12(2-1)\\\\\\\\ &=\dfrac12\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{\pi/6}_{0}\sec(2x)\tan(2x)\,dx\, = \dfrac12$